## Re: [PrimeNumbers] Conjecture Ludovicus V

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• ... Yup: ? lcv(p)=sum(x=1,(p-1)/2,ispseudoprime(x^2+(p-x)^2)) ? lcv(40037) 2641 ... Note that a mean tells you nothing about a limit, in fact, the mean may
Message 1 of 6 , Nov 8, 2012
--- On Thu, 11/8/12, Ludovicus <luiroto@...> wrote:
> "I we decompose any prime p such that
> p = x + y, there are (p-1)/2 forms of doing that. Of those,
> many can produce: x^2 + y^2 = prime.
>
> CONJECTURE. Calling #prime2 that quantity of primes, we have:
> Be R =  #prime2*Log(p) / p . Then Lim. R = 0.70..
> when p--->inf.
>
> Example. p = 40037  ; #prime2 = 2641 ; R = 0.6991

Yup:
? lcv(p)=sum(x=1,(p-1)/2,ispseudoprime(x^2+(p-x)^2))
? lcv(40037)
2641

> From p = 200003 to 100271 the mean R = 0.700205.

Note that a mean tells you nothing about a limit, in fact, the mean may hide the fact that there's no limit, if it fluctuates between a liminf and a limsup. However, the mean is an equally valid value to be interested in, so that's not a problem.

> I will call that conjecture : "Conjecture Ludovicus V"

Not enough data to deny it, and some anecdotal evidence to support it:

? lcv(p)=sum(x=1,(p-1)/2,ispseudoprime(x^2+(p-x)^2))*log(p)/p
time = 0 ms.
? for(n=1,30,p=nextprime(2^n);print(n" "p" "lcv(p)))
1 2 0
2 5 0.6437751649736401498403037333
3 11 0.6539714380359192392896209758
4 17 0.6666384338955802541763610866
5 37 0.7807390081392917717552639289
6 67 0.6903226688552332336771759994
7 131 0.8187354283238575112312009790
8 257 0.7341190151223247982255292090
9 521 0.7204318666126718139833661082
10 1031 0.6258588331848470155706204411
11 2053 0.6910047148395137088493983438
12 4099 0.7082595544903668720682217777
13 8209 0.6982896023307951368923678478
14 16411 0.6978693804400682993622951493
15 32771 0.7097359994440564892776086479
16 65537 0.7004141483596733103866463592
17 131101 0.6948838285968673741945871092
18 262147 0.6944460786798661063812187741
19 524309 0.6969871254592594758140428033
20 1048583 0.6992402515337114838314592173
21 2097169 0.7007186928388776062866463444
22 4194319 0.6970925620012833876499535047
23 8388617 0.6951323183116127789119946883
24 16777259 0.6948711428730801610212743660
25 33554467 0.6956608736980247707005160260
^C

This is probably not the correct way to approach the problem, as it only produces.

A better way might be to split prime p=4n+1 into x^2+y^2 (Cornaccia-Smith, or simply factor in Z[I]), tally x+y, and then for each prime q, once sqrt(2p) exceeds q, read off the value of the tally for q.

This script is a starting point, but doesn't care about the q values, just makes the tallies:

? lim=1000;tally=vectorsmall(lim/2);forprime(p=5,lim*lim/2,if(p%4==1,f=factor(p+0*I)[2,1];p2=real(f)+imag(f);tally[(p2+1)/2]++));tally
time = 626 ms.
Vecsmall([0, 1, 2, 2, 2, 3, 3, 4, 4, 4, 3, 4, 8, 4, 6, 5, 4, 9, 8, 6, 9, 7, 7, 7, 5, 7, 9, 14, 8, 9, 11, 7, 17, 11, 10, 9, 11, 9, 8, 13, 9, 15, 20, 11, 14, 13, 8, 18, 14, 10, 18, 16, 10, 17, 16, 13, 20, 20, 13, 14, 17, 12, 23, 18, 14, 22, 15, 17, 18, 21, 12, 19, 29, 16, 23, 21, 14, 27, 24, 14, 19, 24, 19, 22, 25, 16, 24, 24, 13, 22, 28, 14, 35, 25, 15, 30, 23, 26, 27, 24, 17, 26, 36, 18, 21, 29, 22, 39, 27, 23, 32, 30, 27, 30, 28, 17, 29, 43, 20, 24, 28, 17, 38, 34, 20, 30, 28, 31, 34, 27, 23, 33, 48, 24, 36, 35, 26, 41, 31, 23, 37, 32, 33, 27, 38, 22, 36, 53, 20, 38, 35, 26, 53, 37, 28, 41, 42, 33, 42, 37, 28, 44, 58, 25, 35, 42, 25, 58, 43, 22, 37, 37, 36, 42, 38, 29, 34, 57, 23, 43, 39, 26, 60, 38, 31, 42, 45, 49, 56, 44, 32, 44, 45, 28, 44, 42, 32, 58, 42, 20, 45, 52, 42, 48, 51, 33, 48, 68, 28, 47, 50, 35, 76, 43, 38, 48, 49, 46, 51, 46, 27, 48, 69, 39, 57, 54, 34, 62, 57, 44, 53, 53, 49, 53, 42, 39, 47, 75, 38, 60, 59, 25, 81, 57, 31, 52, 65, 39,
47, 59, 32, 51, 74, 42, 54, 48, 30, 76, 52, 41, 60, 56, 51, 62, 62, 45, 67, 82, 41, 48, 69, 40, 82, 58, 47, 63, 56, 52, 57, 77, 42, 68, 90, 34, 61, 67, 43, 80, 67, 43, 52, 56, 66, 67, 57, 40, 70, 77, 40, 68, 62, 50, 75, 68, 35, 77, 71, 60, 66, 67, 45, 55, 90, 38, 77, 69, 48, 90, 69, 45, 68, 68, 62, 74, 66, 32, 72, 88, 49, 72, 67, 64, 82, 73, 51, 66, 65, 62, 76, 63, 49, 76, 98, 55, 86, 74, 38, 105, 81, 48, 70, 77, 72, 61, 83, 56, 69, 94, 54, 74, 56, 51, 110, 75, 51, 72, 72, 53, 67, 70, 55, 84, 103, 45, 59, 73, 46, 88, 72, 33, 68, 61, 61, 79, 66, 53, 71, 92, 39, 67, 69, 42, 84, 63, 38, 48, 72, 64, 70, 67, 42, 61, 73, 44, 61, 67, 42, 79, 54, 39, 63, 63, 46, 61, 62, 44, 59, 73, 39, 61, 60, 35, 78, 41, 35, 55, 53, 45, 54, 51, 31, 43, 66, 40, 47, 59, 41, 65, 50, 39, 54, 46, 43, 45, 39, 31, 42, 62, 35, 41, 50, 26, 56, 46, 29, 45, 34, 40, 35, 38, 32, 42, 48, 23, 45, 29, 28, 50, 39, 21, 37, 30, 27, 29, 40, 24, 29, 35, 21, 25, 29, 18, 42, 30, 16, 22, 23, 22, 20,
20, 14, 19, 21, 12, 15, 14, 10, 14, 10, 2])

YMMV...

Phil
• ... Looks like an expensive method of approximating log(2) ? David (with no sound basis for this guess)
Message 2 of 6 , Nov 11, 2012
Phil Carmody <thefatphil@...> wrote:

> 17 131101 0.6948838285968673741945871092
> 18 262147 0.6944460786798661063812187741
> 19 524309 0.6969871254592594758140428033
> 20 1048583 0.6992402515337114838314592173
> 21 2097169 0.7007186928388776062866463444
> 22 4194319 0.6970925620012833876499535047
> 23 8388617 0.6951323183116127789119946883
> 24 16777259 0.6948711428730801610212743660
> 25 33554467 0.6956608736980247707005160260

Looks like an expensive method of approximating log(2) ?

David (with no sound basis for this guess)
• ... hmmmm you mean ln(2) :-) ... Yann (who wonders what kind of relationship there can be between ln(2) and the threshold voltage of a bipolar transistor...
Message 3 of 6 , Nov 11, 2012
Le 2012-11-11 13:11, djbroadhurst a écrit :
>> 25 33554467 0.6956608736980247707005160260
> Looks like an expensive method of approximating log(2) ?

hmmmm you mean ln(2) :-)

> David (with no sound basis for this guess)

Yann (who wonders what kind of relationship
there can be between ln(2) and the threshold
voltage of a bipolar transistor...
damn coincidences)
• ... ln(2) is just a fussy way of writing log(2) for the benefit people who think that God has 10 fingers. In fact She has exp(1) fingers :-) David
Message 4 of 6 , Nov 11, 2012

> hmmmm you mean ln(2) :-)

ln(2) is just a fussy way of writing log(2) for the
benefit people who think that God has 10 fingers.
In fact She has exp(1) fingers :-)

David
• ... That value went through my head too, certainly. If it is log(2), I would hope that there is a sound reason for it to be so. Phil
Message 5 of 6 , Nov 11, 2012
> --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
> > 17 131101 0.6948838285968673741945871092
> > 18 262147 0.6944460786798661063812187741
> > 19 524309 0.6969871254592594758140428033
> > 20 1048583 0.6992402515337114838314592173
> > 21 2097169 0.7007186928388776062866463444
> > 22 4194319 0.6970925620012833876499535047
> > 23 8388617 0.6951323183116127789119946883
> > 24 16777259 0.6948711428730801610212743660
> > 25 33554467 0.6956608736980247707005160260
>
> Looks like an expensive method of approximating log(2) ?
>
> David (with no sound basis for this guess)

That value went through my head too, certainly. If it is log(2), I would hope that there is a sound reason for it to be so.

Phil
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